1. John E. McMurry Robert C. Fay C H E M I S T R Y Sixth Edition © 2012 Pearson Education, Inc.

2. John E. McMurry Robert C. Fay C H E M I S T R Y Sixth Edition © 2012 Pearson Education, Inc. Unit 6 Applications of Aqueous Equilibria Ebook Ch15

3. OBJECTIVE Learning Objective  To solve problems pertaining to chemical equilibrium Specific Objectives Neutralization pH titration curves Solubility

4. Chapter 15/4 © 2012 Pearson Education, Inc. 6.1 Neutralization Reactions Assume complete dissociation: Strong Acid-Strong Base After neutralization: pH = 7 H 2 O(l) +NaOH(aq)HCl(aq)+NaCl(aq) 2H 2 O(l)H 3 O + (aq) + OH – (aq) (net ionic equation) Neutral

5. Chapter 15/5 © 2012 Pearson Education, Inc. 6.1 Neutralization Reactions H 2 O(l) + CH 3 CO 2 – (aq)CH 3 CO 2 H(aq) + OH – (aq) Assume complete dissociation: Weak Acid-Strong Base After neutralization: pH > 7 (net ionic equation) H 2 O(l) +CH 3 CO 2 H(aq) +NaOH(aq)NaCH 3 CO 2 (aq) Weak base

6. Chapter 15/6 © 2012 Pearson Education, Inc. Assume complete dissociation: Strong Acid-Weak Base After neutralization: pH < 7 NH 4 Cl(aq)HCl(aq)+ NH 3 (aq) H 2 O(l) +H 3 O + (aq) + NH 3 (aq)NH 4 + (aq) (net ionic equation) Weak acid 6.1 Neutralization Reactions

7. Chapter 15/7 © 2012 Pearson Education, Inc. NH 4 + (aq) + CH 3 CO 2 – (aq)CH 3 CO 2 H(aq) + NH 3 (aq) Weak Acid-Weak Base After neutralization: pH = ?, we need more information! NH 4 CH 3 CO 2 (aq)CH 3 CO 2 H(aq) + NH 3 (aq) (net ionic equation) Assume complete dissociation: 6.1 Neutralization Reactions Weak base Weak acid Since both the acid and the base are weak, it complicates the titration. Typical general chemistry titrations are not done with both of them weak. The titration generally does not go to 100%. The pH of the solution can be acidic or basic and depends on the weak acid and weak base.

8. Chapter 15/8 © 2012 Pearson Education, Inc. 6.1 Neutralization Reactions Application 1 Write a balanced net ionic equation for the neutralization of equal molar amounts of nitric acid and methylamine (CH 3 NH 2 ). Indicate whether the pH after neutralization is greater than, equal to, or less than 7.

9. Chapter 15/9 © 2012 Pearson Education, Inc. 6.1 Neutralization Reactions Strategy: The formulas that should appear in the net ionic equation depend on whether the acid and base are strong (completely dissociated) or weak (largely undissociated). The pH after neutralization depends on the acid–base properties of the cation and anion in the resulting salt solution (Section 14.14). Solution: Because HNO 3 is a strong acid and CH 3 NH 2 is a weak base, the neutralization proceeds to completion and the net ionic equation is After neutralization, the solution contains CH 3 NH 3 +, a weak acid, and NO 3 –, which has no acidic or basic properties. Therefore, the pH is less than 7.

10. Chapter 15/10 © 2012 Pearson Education, Inc. 6.2 The Common-Ion Effect: Le Chatelier Principle’s H 3 O + (aq) + CH 3 CO 2 – (aq)CH 3 CO 2 H(aq) + H 2 O(l) Common-Ion Effect: The shift in the position of an equilibrium on addition of a substance that provides an ion in common with one of the ions already involved in the equilibrium Adding CH 3 CO 2 – or H 3 O + move the equilibrium to the left, the solution is less acid and the pH increase Adding CH 3 CO 2 H or H 2 O move the equilibrium to the right, the soluion is more acid and the pH decrease

11. Chapter 15/11 © 2012 Pearson Education, Inc. The pH of 0.10 M acetic acid is 2.89. Calculate the pH of a solution that is prepared by dissolved 0.10 mol of acetic acid and 0.10 mol sodium acetate in enough water to make 1.00 L of solution. 6.2 The Common-Ion Effect: Le Chatelier Principle’s Application 2

12. Chapter 15/12 © 2012 Pearson Education, Inc. H 3 O + (aq) + CH 3 CO 2 – (aq)CH 3 CO 2 H(aq) + H 2 O(l) 0.10≈00.10 –x–x +x 0.10 – x x0.10 + x [H 3 O + ][CH 3 CO 2 – ] [CH 3 CO 2 H] K a = 6.2 The Common-Ion Effect: Le Chatelier Principle’s Solution

13. Chapter 15/13 © 2012 Pearson Education, Inc. x(0.10) 0.10 ≈1.8 × 10 –5 = (x)(0.10 + x) (0.10 – x) pH = – log([H 3 O + ]) = – log(1.8 × 10 –5 ) = 4.74 x = [H 3 O + ] = 1.8 × 10 –5 M 6.2 The Common-Ion Effect: Le Chatelier Principle’s The addition of acetate ion to a solution of acetic acid suppresses the dissociation of the acid. The equilibrium shifts to the left, then the pH increase

14. Chapter 15/14 © 2012 Pearson Education, Inc. 6.2 The Common-Ion Effect: Le Chatelier Principle’s

15. Chapter 15/15 © 2012 Pearson Education, Inc. 6.2 The Common-Ion Effect: Le Chatelier Principle’s

16. Chapter 15/16 © 2012 Pearson Education, Inc. 6.3 Buffer Solutions Buffer Solution: A solution which contains a weak acid and its conjugate base and resists drastic changes in pH a) CH 3 CO 2 H + CH 3 CO 2 – b) HF + F – c) NH 4 + + NH 3 d) H 2 PO 4 – + HPO 4 2– Weak acid + Conjugate base For example:

17. Chapter 15/17 © 2012 Pearson Education, Inc. Conjugate base (NaCH 3 CO 2 ) Weak acid H 3 O + (aq) + CH 3 CO 2 – (aq)CH 3 CO 2 H(aq) + H 2 O(l) H 2 O(l) + CH 3 CO 2 H(aq)CH 3 CO 2 – (aq) + H 3 O + (aq) 100% Addition of H 3 O 1+ to a buffer: Addition of OH 1– to a buffer: H 2 O(l) + CH 3 CO 2 – (aq)CH 3 CO 2 H(aq) + OH – (aq) 100% 6.3 Buffer Solutions

19. Chapter 15/19 © 2012 Pearson Education, Inc. 6.3 Buffer Solutions

20. Chapter 15/20 © 2012 Pearson Education, Inc. 6.4 The Henderson- Hasselbalch Equation Conjugate baseWeak acid H 3 O + (aq) + Base(aq)Acid(aq) + H 2 O(l) [Acid] [Base] [H 3 O + ] = K a K a = [H 3 O + ][Base] [Acid] H 3 O + (aq) + CH 3 CO 2 – (aq)CH 3 CO 2 H(aq) + H 2 O(l)

21. Chapter 15/21 © 2012 Pearson Education, Inc. 6.4 The Henderson- Hasselbalch Equation [Base] [Acid] pH = pK a + log [Acid] [Base] [H 3 O + ] = K a –log([H 3 O + ]) = –log(K a ) – log [Acid] [Base]

22. Chapter 15/22 © 2012 Pearson Education, Inc. 6.3 Buffer Solutions (a)Use the Henderson–Hasselbalch equation to calculate the pH of a buffer solution that is 0.45 M in NH 4 Cl and 0.15 M in NH 3. (b) How would you prepare an NH 4 Cl-NH 3 buffer that has a pH of 9.00? Application 3

23. Chapter 15/23 © 2012 Pearson Education, Inc. 6.3 Buffer Solutions Strategy (a)Since NH 4 + is the weak acid in an NH 4 –NH 3 buffer solution, we need to find the pK a for NH 4 + from the tabulated Kb value for NH 3 (Appendix C). Then, substitute [NH 3 ], [NH 4 + ], and the pK a value into the Henderson–Hasselbalch equation to find the pH. (b)Use the Henderson–Hasselbalch equation to calculate the [NH 3 ]/[NH 4 +] ratio from the desired pH and the pK a value for NH 4 +.

24. Chapter 15/24 © 2012 Pearson Education, Inc. 6.3 Buffer Solutions Solution (a) Since K b for NH 3 is 1.8  10 -5, K a for ist NH 4 + is 5.6  10 –10 and pK a = 9.25. pK a = -log K a = –log (5.6  10 -10 ) = 9.25 Since [base] = [NH 3 ] = 0.15 M and [acid] = [NH 4 + ] = 0.45 M, The pH of the buffer solution is 8.77.

25. Chapter 15/25 © 2012 Pearson Education, Inc. 6.3 Buffer Solutions (b) Rearrange the Henderson–Hasselbalch equation to obtain an expression for the relative amounts of NH 3 and NH 4 + in a solution having pH = 9.00: Therefore, The solution must contain 0.56 mol of NH 3 for every 1.00 mol of NH 4 Cl, but the volume of the solution isn’t critical. One way of preparing the buffer would be to combine 1.00 mol of NH 4 Cl (53.5 g) with 0.56 mol of NH 3 (say, 560 mL of 1.00 M NH 3 ).

26. Chapter 15/26 © 2012 Pearson Education, Inc. 6.5 pH Titration Curves Titration: A procedure for determining the concentration of a solution by allowing a carefully measured volume to react with a solution of another substance (the standard solution) whose concentration is known Equivalence Point: The point at which stoichiometrically equivalent quantities of acid and base have been mixed together

27. Chapter 15/27 © 2012 Pearson Education, Inc. 6.5 pH Titration Curves

28. 2H 2 O(l)H 3 O + (aq) + OH – (aq) 100% 6.5 pH Titration Curves Strong acid - Strong base HCl(aq) + NaOH(aq)NaCl(aq) + H 2 O(l) 100% - Molecular equation - Net ionic equation

29. Chapter 15/29 © 2012 Pearson Education, Inc. 1. Before Addition of Any NaOH There is only HCl present in the solution. For 0.100 M HCl, pH = 1 6.5 pH Titration Curves Strong acid - Strong base

30. Chapter 15/30 2. Before the Equivalence Point The majority of H 3 O + reacts with NaOH but there is still some H 3 O + left. We are close to the equivalent point and pH < 7 6.5 pH Titration Curves Strong acid - Strong base

31. Chapter 15/31 3.At the Equivalence Point All H 3 O + reacts with the quantity of added OH –. The products in solution are NaCl(aq) and H 2 O(l). There is no acid and no base left. pH = 7 6.5 pH Titration Curves Strong acid - Strong base

32. Chapter 15/32 4.Beyond the Equivalence Point There is no H 3 O + left, the added OH – does not reacts and remain in solution. The pH > 7 and continue to increase as we add more OH – 6.5 pH Titration Curves Strong acid - Strong base

33. 6.5 pH Titration Curves Strong acid - Strong base

34. H 2 O(l) + CH 3 CO 2 – (aq)CH 3 CO 2 H(aq) + OH – (aq) 6.5 pH Titration Curves Weak acid - Strong base

36. 6.5 pH Titration Curves Weak acid - Strong base

37. Chapter 15/37 © 2012 Pearson Education, Inc. 6.5 pH Titration Curves Strong acid - Weak base HCl(aq) + NH 3 (aq)NH 4 Cl(aq) + H 2 O(l) 100% - Molecular equation - Net ionic equation NH 4 + (aq) + H 2 O(l)NH 3 (aq) + H 3 O + (aq) 100%

38. Chapter 15/38 © 2012 Pearson Education, Inc. 1.Before Addition of Any HCl Only the weak base 0.100 M NH 3 is present in the solution. pH > 7 6.5 pH Titration Curves Strong acid - Weak base

39. Chapter 15/39 © 2012 Pearson Education, Inc. 2.Before the Equivalence Point Half of the initial NH 3 reacts with acid and form NH 4 +. What is left in the solution is half of NH 3 and the same quantity of NH 4 + is forms: This is a buffer solution. pH = pKa NH 4 + (aq) + H 2 O(l)NH 3 (aq) + H 3 O + (aq) 100% 6.5 pH Titration Curves Strong acid - Weak base

40. Chapter 15/40 © 2012 Pearson Education, Inc. 3.At the Equivalence Point H 3 O + (aq) + NH 3 (aq)NH 4 + (aq) + H 2 O(l) There is no NH 3 left, all is transform into the weak acid NH 4 +. pH < 7 6.5 pH Titration Curves Strong acid - Weak base

41. Chapter 15/41 © 2012 Pearson Education, Inc. 4.Beyond the Equivalence Point More strong acid H 3 O + is added in the solution and there is no reaction. The solution become more acid and the pH decrease, pH < 7 6.5 pH Titration Curves Strong acid - Weak base

42. Chapter 15/42 © 2012 Pearson Education, Inc. 6.5 pH Titration Curves Strong acid - Weak base Application 4 a)What is the pH at the equivalence point of a weak base-strong acid titration if 20.00 mL of NaOCl requires 28.30 mL of 0. 50 M HCl? Ka = 3.0 × 10 -8 for HOCl. Ans: 4.03 b)What is the pH if 14.15 mL of 0. 50 M HCl was used? What is the characterictic of this point? Ans: 7.52 c)What is the pH if 20 mL of 0. 50 M HCl was used? What is the characterictic of this point? Ans: 7.14 d)What is the pH if 35 mL of 0. 50 M HCl was used? What is the characterictic of this point? Ans: 1.22

43. Polyprotic Acid-Strong Base Titrations

44. Chapter 15/44 © 2012 Pearson Education, Inc. Solubility Equilibria Ca 2+ (aq) + 2F – (aq)CaF 2 (s) K sp = [Ca 2+ ][F – ] 2 mM n+ (aq) + xX y– (aq)MmXx(s)MmXx(s) K sp = [M n+ ] m [X y– ] x

45. Chapter 15/45 © 2012 Pearson Education, Inc. 6.6 Measuring K sp and Calculating Solubility from K sp If the concentrations of Ca 2+ (aq) and F 1– (aq) in a saturated solution of calcium fluoride are known, K sp may be calculated. Ca 2+ (aq) + 2F – (aq)CaF 2 (s) K sp = [Ca 2+ ][F – ] 2 = (2.0 × 10 –4 )(4.1 × 10 –4 ) 2 = 3.4 x 10 –11 [F – ] = 4.1 × 10 –4 M[Ca 2+ ] = 2.0 × 10 –4 M (at 25 o C)

46. Chapter 15/46 © 2012 Pearson Education, Inc. 6.6 Measuring K sp and Calculating Solubility from K sp

47. Chapter 15/47 © 2012 Pearson Education, Inc. 6.6 Measuring K sp and Calculating Solubility from K sp a)Calculate the molar solubility of MgF 2 in water at 25 o C. K sp (MgF 2 (s))= 7.4 × 10 – 11 Mg 2+ (aq) + 2F – (aq)MgF 2 (s) Application 5 b) Calculate the molar solubility of MgF 2 in 0.10 M NaF at 25 o C. K sp (MgF 2 (s))= 7.4 × 10 – 11 c) Why does the solubility decrease in the presence of NaF, with a common ion F - ?

48. Chapter 15/48 © 2012 Pearson Education, Inc. 6.6 Measuring K sp and Calculating Solubility from K sp x = [Mg 2+ ] = Molar solubility = 2.6 × 10 – 4 M K sp = 7.4 × 10 – 11 = [Mg 2+ ][F – ] 2 = (x)(2x) 2 4x 3 = 7.4 × 10 – 11 Mg 2+ (aq) + 2F – (aq)MgF 2 (s) x2x Solution a)

49. Chapter 15/49 © 2012 Pearson Education, Inc. Solution 7.4 × 10 – 11 = (x)(0.10 + 2x) 2 ≈ (x)(0.10) 2 K sp = 7.4 × 10 – 11 = [Mg 2+ ][F – ] 2 = (x)(0.10 + 2x) 2 = 7.4 × 10 – 9 M (0.10) 2 7.4 × 10 – 9 x = [Mg 2+ ] = Molar solubility = Mg 2+ (aq) + 2F – (aq)MgF 2 (s) x0.10 + 2x b) 6.6 Measuring K sp and Calculating Solubility from K sp

50. Chapter 15/50 © 2012 Pearson Education, Inc. Mg 2+ (aq) + 2F – (aq)MgF 2 (s) Solubility and the Common-Ion Effect Molar solubility in 0.10 M NaF: Molar solubility without NaF: 7.4 × 10 – 9 M 2.6 × 10 – 4 M c) Answer: Le Châtelier’s Principle. By adding NaF, we increase the amount of F - ion in the solution, this favor the shift of the equilibrium to the left thus decrease the solubility. 6.6 Measuring K sp and Calculating Solubility from K sp

51. Mg 2+ (aq) + 2F – (aq)MgF 2 (s) Solubility and the Common-Ion Effect 6.7 Factors That Affect Solubility

52. Solubility and the pH of the Solution Ca 2+ (aq) + HCO 3 – (aq) + H 2 O(l)CaCO 3 (s) + H 3 O + (aq) 6.7 Factors That Affect Solubility

53. Chapter 15/53 © 2012 Pearson Education, Inc. Ag(NH 3 ) 2 + (aq)Ag(NH 3 ) + (aq) + NH 3 (aq) Ag(NH 3 ) 2 + (aq)Ag + (aq) + 2NH 3 (aq) K f = 1.7 × 10 7 K 1 = 2.1 × 10 3 Ag(NH 3 ) + (aq)Ag + (aq) + NH 3 (aq) K 2 = 8.1 × 10 3 K f is the formation constant. How is it calculated? (at 25 o C) Solubility and the Formation of Complex Ions 6.7 Factors That Affect Solubility The complex is form from the reaction with the reactant NH3, thus favor the shift of the equilibrium to the right and increase the solbility

54. Chapter 15/54 © 2012 Pearson Education, Inc. = (2.1 × 10 3 )(8.1 × 10 3 ) = 1.7 × 10 7 [Ag(NH 3 ) 2 + ] [Ag + ][NH 3 ] 2 K 1 = [Ag(NH 3 ) + ] [Ag + ][NH 3 ] K 2 = [Ag(NH 3 ) 2 + ] [Ag(NH 3 ) + ][NH 3 ] K f = K 1 K 2 = Solubility and the Formation of Complex Ions 6.7 Factors That Affect Solubility Ag(NH 3 ) 2 + (aq)Ag + (aq) + 2NH 3 (aq) KfKf

56. Chapter 15/56 © 2012 Pearson Education, Inc. Ag(NH 3 ) 2 + (aq)Ag + (aq) + 2NH 3 (aq) Solubility and the Formation of Complex Ions 6.7 Factors That Affect Solubility

57. Chapter 15/57 © 2012 Pearson Education, Inc. Solubility and Amphoterism Al(OH) 4 – (aq)Al(OH) 3 (s) + OH – (aq) Al 3+ (aq) + 6H 2 O(l)Al(OH) 3 (s) + 3H 3 O + (aq) In base: In acid: Aluminum hydroxide Al(OH) 3 is soluble both in strongly acidic and in strongly basic solutions. 6.7 Factors That Affect Solubility amphoteric compound: is a molecule or ion that can react both as an acid as well as a base.

58. Chapter 15/58 © 2012 Pearson Education, Inc. Solubility and Amphoterism 6.7 Factors That Affect Solubility Example of Aluminum hydroxide Al(OH) 3

59. Solubility and Amphoterism 6.7 Factors That Affect Solubility

60. Chapter 15/60 © 2012 Pearson Education, Inc. Ca 2+ (aq) + 2F – (aq)CaF 2 (s) Q c = [Ca 2+ ] t [F – ] t 2 K sp = [Ca 2+ ][F – ] 2 Q c is called the ion-product (IP). 6.8 Separation of Ions by Selective Precipitation Q c (IP) use initial concentrations of ions, mention t K sp use concentrations of ions at the equilibrium

61. Chapter 15/61 © 2012 Pearson Education, Inc. If the value of IP is greater than k sp, then there is more ions in solution than what it should be at the equilibrium: the solution is supersaturated. Following Le Chatelier principle’s, the amount of ions should decrease, this will favor the formation of the precipitate, thus reducing the ions concentrations until IP equals k sp. At that point solubility equilibrium is reached and the solution is saturated. Compare the IP to K sp 6.8 Separation of Ions by Selective Precipitation

62. Chapter 15/62 © 2012 Pearson Education, Inc. The solution is supersaturated and precipitation will occur. The solution is saturated and equilibrium exists already. The solution is unsaturated and precipitation will not occur. If IP > K sp : If IP = K sp : If IP < K sp : Compare the IP to K sp 6.8 Separation of Ions by Selective Precipitation

63. M 2+ (aq) + H 2 S(aq) + 2H 2 O(l)MS(s) + 2H 3 O + (aq) K spa = [M 2+ ][H 2 S] [H 3 O + ] 2 When an acid is use to dissolve a precipitate, then the concentration of the acid will appear as reactant in the equilibirum constant

64. Chapter 15/64 © 2012 Pearson Education, Inc. Q c = [M 2+ ] t [H 2 S] t [H 3 O + ] t 2 Adjusting the pH changes Q c (ion-product). M 2+ (aq) + H 2 S(aq) + 2H 2 O(l)MS(s) + 2H 3 O + (aq) K spa = [M 2+ ][H 2 S] [H 3 O + ] 2 6.8 Separation of Ions by Selective Precipitation

65. Chapter 15/65 © 2012 Pearson Education, Inc. 6.8 Separation of Ions by Selective Precipitation Application 6 Will a precipitate form when 0.150 L of 0.10 M Pb(NO 3 ) 2 and 0.100 L of 0.20 M NaCl are mixed?

66. Chapter 15/66 © 2012 Pearson Education, Inc. 6.8 Separation of Ions by Selective Precipitation Since ionic compounds are strong electrolytes, Pb(NO 3 ) 2 and NaCl exist in solution as separate cations and anions. Use the solubility guidelines discussed in Section 4.4 to decide which ions might form a precipitate, and calculate their concentrations after mixing. Then calculate the IP for the possible precipitate and compare it with the value of K sp. Strategy

67. Chapter 15/67 © 2012 Pearson Education, Inc. 6.8 Separation of Ions by Selective Precipitation After the two solutions are mixed, the combined solution contains Pb 2+, NO 3 , Na +, and Cl  ions and has a volume of 0.150 L + 0.100 L = 0.250 L. Because sodium salts and nitrate salts are soluble in water, the only compound that might precipitate is PbCl 2, which has K sp = 1.2  10  5 (Appendix C). To calculate the value of IP for PbCl 2, first calculate the number of moles of Pb 2+ and Cl  in the combined solution: Moles Pb 2+ = (0.150 L)(0.10 mol/L) = 1.5  10  2 mol Moles Cl  = (0.100 L)(0.20 mol/L) = 2.0  10  2 mol Solution

68. Chapter 15/68 © 2012 Pearson Education, Inc. 6.8 Separation of Ions by Selective Precipitation Then convert moles to molar concentrations: The ion product is IP = [Pb 2+ ] t [Cl  ] t 2 = (6.0  10  2 )(8.0  10  2 ) 2 = 3.8  10  4 Since K sp = 1.2  10  5, IP is greater than K sp and PbCl 2 will therefore precipitate.

69. 6.9 Qualitative Analysis See in this graph the selectivity of some substances that forms a precipitates with specific ions

70. Chapter 15/70 © 2012 Pearson Education, Inc. 6.9 Qualitative Analysis

71. Chapter 15/71 © 2012 Pearson Education, Inc. 6.9 Qualitative Analysis Application 7 How will you separate a mixture of Ag +, Al 3+ and Ca 2+ ?

72. Chapter 15/72 © 2012 Pearson Education, Inc. 6.9 Qualitative Analysis Solution Ag + (aq) Al 3+ (aq) Ca 2+ (aq) NaCl AgCl(s) Al 3+ (aq) Ca 2+ (aq) Na + (aq) filtration Al 3+ (aq) Ca 2+ (aq) Na + (aq) AgCl(s) NaOH Al(OH) 3 (s) Ca 2+ (aq) Na + (aq) filtration Al(OH) 3 (s) Ca 2+ (aq) Na + (aq) Na 2 CO 3 filtration CaCO 3 (s) Na + (aq)